Chapter 5, Problem 5G.4BST

Interpretation Introduction

Interpretation:

The value of ${\text{ΔG}}_{\text{r}}$ at $\text{298}\text{K}$ for the given reaction has to be calculated.  Also the spontaneous direction of the reaction has to be given.

  ${\text{N}}_2{\text{O}}_{\text{4}\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NO}}_{2\left(\text{g}\right)}{\text{ΔG}}_{\text{r}}^{\text{°}}\text{=}\text{+4}\text{.73}\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

Concept Introduction:

The Gibb’s free energy of a reaction can be calculated by the expression,

  $\begin{array}{l}{\text{ΔG}}_{\text{r}}\text{=}{\text{ΔG}}_{\text{r}}^{\text{°}}\text{+}\text{RT}\text{ln}\text{Q}\\ \\ \text{where,}\\ {\text{ΔG}}_{\text{r}}\text{=}\text{Gibb's}\text{free}\text{energy}\\ {\text{ΔG}}_{\text{r}}^{\text{°}}\text{=}\text{Standard}\text{Gibb's}\text{free}\text{energy}\\ \text{Q}\text{=}\text{Reaction}\text{quotient}\end{array}$

The value of reaction quotient can be calculated by the expression,

  $\text{Q}\text{=}\frac{{\left({\text{a}}_{\text{C}}\right)}^{\text{c}}{\left({\text{a}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{a}}_{\text{A}}\right)}^{\text{a}}{\left({\text{a}}_{\text{B}}\right)}^{\text{b}}}$